组合数学。。。(每做一题都是这么艰难)
Round Numbers
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 7607 Accepted: 2615

Description

The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets to be milked first. They can't even flip a coin because it's so hard to toss using hooves.

They have thus resorted to "round number" matching. The first cow picks an integer less than two billion. The second cow does the same. If the numbers are both "round numbers", the first cow wins,
otherwise the second cow wins.

A positive integer N is said to be a "round number" if the binary representation of N has as many or more zeroes than it has ones. For example, the integer 9, when written in binary form, is 1001. 1001 has two zeroes and two ones; thus, 9 is a round number. The integer 26 is 11010 in binary; since it has two zeroes and three ones, it is not a round number.

Obviously, it takes cows a while to convert numbers to binary, so the winner takes a while to determine. Bessie wants to cheat and thinks she can do that if she knows how many "round numbers" are in a given range.

Help her by writing a program that tells how many round numbers appear in the inclusive range given by the input (1 ≤ Start < Finish ≤ 2,000,000,000).

Input

Line 1: Two space-separated integers, respectively Start and Finish.

Output

Line 1: A single integer that is the count of round numbers in the inclusive range Start..Finish

Sample Input

2 12

Sample Output

6

Source

USACO 2006 November Silver

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int c[35][35],bit[35];

void Init()
{
    for(int i=0;i<=33;i++)
        c=c[0]=1;
    for(int i=2;i<=33;i++)
        for(int j=1;j<i;j++)
            c[j]=c[i-1][j-1]+c[i-1][j];
}

int calu(int x)
{
    if(x<=1) return 0;
    memset(bit,0,sizeof(bit));
    int pos=32,ret=0;
    for(int i=0;i<32;i++)
        if(x&(1<<i)) bit=1;
    for(;bit[pos]==0&&pos>=0;pos--);
    ///n-1....1
    for(int i=pos;i>=1;i--)
    {
        int k=i-1;
        if(i%2)
            ret+=((1<<k)-c[k][k>>1])>>1;
        else
            ret+=(1<<k)>>1;
    }
    ///N
    int n1=0,n0=0;
    for(int i=pos;i>=0;i--)
    {
        if(bit) n1++;
        else n0++;
    }
    if(n0>=n1) ret++;
    n1=1,n0=0;
    for(int i=pos-1;i>=0;i--)
    {
        if(bit)
        {
            for(int j=i;j>=0&&n0+j+1>=n1+i-j;j--)
            {
                ret+=c[j];
            }
            n1++;
        }
        else n0++;
    }
    return ret;
}

int main()
{
    Init();
    int a,b;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        printf("%d\n",calu(b)-calu(a-1));
    }
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )

POJ 3252 Round Numbers的更多相关文章

  1. POJ 3252 Round Numbers(组合)

    题目链接:http://poj.org/problem?id=3252 题意: 一个数的二进制表示中0的个数大于等于1的个数则称作Round Numbers.求区间[L,R]内的 Round Numb ...

  2. [ACM] POJ 3252 Round Numbers (的范围内的二元0数大于或等于1数的数目,组合)

    Round Numbers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8590   Accepted: 3003 Des ...

  3. poj 3252 Round Numbers(数位dp 处理前导零)

    Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, P ...

  4. POJ 3252 Round Numbers 数学题解

    Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, P ...

  5. POJ 3252 Round Numbers 组合数学

    Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13381   Accepted: 5208 Description The ...

  6. POJ 3252 Round Numbers(组合数学)

    Round Numbers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10223   Accepted: 3726 De ...

  7. POJ 3252 Round Numbers(数位dp&amp;记忆化搜索)

    题目链接:[kuangbin带你飞]专题十五 数位DP E - Round Numbers 题意 给定区间.求转化为二进制后当中0比1多或相等的数字的个数. 思路 将数字转化为二进制进行数位dp,由于 ...

  8. POJ - 3252 - Round Numbers(数位DP)

    链接: https://vjudge.net/problem/POJ-3252 题意: The cows, as you know, have no fingers or thumbs and thu ...

  9. poj 3252 Round Numbers 【推导·排列组合】

    以sample为例子 [2,12]区间的RoundNumbers(简称RN)个数:Rn[2,12]=Rn[0,12]-Rn[0,1] 即:Rn[start,finish]=Rn[0,finish]-R ...

随机推荐

  1. 偶然发现的Unity3d,两点之间的距离计算。

    无意间查了一下Vector3的API,发现了一个方法. magnitude  Returen the length of vector(Read Only). 然后就试了一下这个方法. Vector3 ...

  2. 数据结构作业——brothers(二叉树)

    brothers Description 给你一棵节点编号从 1 到 n 的,根节点为 1 的二叉树.然后有 q 个询问,每个询问给出一个整数表示树的节点,要求这个节点的兄弟节点数目和堂兄弟节点的数目 ...

  3. 【Alpha版本】 第七天 11.15

    一.站立式会议照片: 二.项目燃尽图: 三.项目进展: 成 员 昨天完成任务 今天完成任务 明天要做任务 问题困难 心得体会 胡泽善 完成我要招聘的招聘详情显示,完成简历填写及显示功能 完成我要应聘的 ...

  4. Linux 中强大且常用命令:find、grep

    在linux下面工作,有些命令能够大大提高效率.本文就向大家介绍find.grep命令,他哥俩可以算是必会的linux命令,我几乎每天都要用到他们.本文结构如下:    find命令        f ...

  5. jsp分页功能

    http://blog.csdn.net/xiazdong/article/details/6857515

  6. sql between and

    sql中的 a between 'a' and 'b' 基本上是代表 'a'>=a and 'b'<=a

  7. 将字符串转化为数字(Convert和Parse的用法)

    字符串必须是数字,不要超过转换成目标数字类型的范围.超过的话系统也会报错(溢出). static void Main(string[] args) { string s; int i; Console ...

  8. curl方式创建elasticsearch的mapping

    curl方式创建elasticsearch的mapping curl -XPUT 'http://192.168.1.105:9200/bank/item2/_mapping' -d '{ " ...

  9. 从零开始HTML(一 2016/10/17)

    就是准备跟着霹雳猿教程过一遍HTML啦,边看边记录更便于理解记忆吧~ 1.属性 HTML 标签可以拥有属性.属性提供了有关 HTML 元素的更多的信息.属性总是以名称/值对的形式出现,比如:name= ...

  10. Codeforces Problem 708A Letters Cyclic Shift

     题目链接: http://codeforces.com/problemset/problem/708/A 题目大意: 从字符串s中挑选出一个子串(非空),将该子串中的每个字母均替换成前一个字母,如' ...